2.12.1.5 Example 5 \(y^{\prime }=x\left ( 1+\frac {2y}{x}+\frac {y^{2}}{x^{4}}\right ) \)

Solve

\[ y^{\prime }=x\left ( 1+\frac {2y}{x}+\frac {y^{2}}{x^{4}}\right ) \]

The first step is to identify if this is class G and find \(F\). We start by multiplying the RHS by \(\frac {x}{y}\) (regardless of what is in the RHS) which gives

\begin{align*} y^{\prime } & =\frac {x}{y}\left ( x\left ( 1+\frac {2y}{x}+\frac {y^{2}}{x^{4}}\right ) \right ) \\ & =\frac {x^{2}}{y}+2x+\frac {y}{x^{2}}\\ & =\frac {\left ( x^{2}+y\right ) ^{2}}{x^{2}y}\\ & =F\left ( x,y\right ) \end{align*}

Next we check if \(F\left ( x,y\right ) \) has \(y\) or not in it. If so, then let the RHS above be \(F\left ( x,y\right ) \) and now do

\begin{align*} f_{x} & =x\frac {\partial F}{\partial x}\\ & =\frac {2x^{4}-2y^{2}+2x^{3}y}{x^{2}y}\end{align*}

And let

\begin{align*} f_{y} & =y\frac {\partial F}{\partial y}\\ & =\frac {y^{2}-x^{4}}{x^{2}y}\end{align*}

Now we check, if \(f_{y}=0\) then this is not Homogeneous type G. Else we now need to determine value of \(\alpha \). This is done as follows.

\begin{align*} \alpha & =\frac {fx}{f_{y}}\\ & =\frac {\frac {2x^{4}-2y^{2}+2x^{3}y}{x^{2}y}}{\frac {y^{2}-x^{4}}{x^{2}y}}\\ & =\frac {2x^{4}-2y^{2}+2x^{3}y}{y^{2}-x^{4}}\end{align*}

Since \(\alpha \) has \(x\) and \(y\) then this is not Homogeneous type G.